1. Lurking since 1997
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## Re: Do you know the correct answer?

Originally Posted by Paul Pless
544 replies!
449 replies. 95 posts were made by moi.

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## Re: Do you know the correct answer?

Choosing the door that Monty didn't open is identical to having chosen both the doors you didn't choose. You win if either one has the car, regardless of the fact that Monty shows you the one that doesn't. That's 66% odds.

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## Re: Do you know the correct answer?

Originally Posted by George.
Choosing the door that Monty didn't open is identical to having chosen both the doors you didn't choose. You win if either one has the car, regardless of the fact that Monty shows you the one that doesn't. That's 66% odds.
Another clear way of stating the situation.

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## Re: Do you know the correct answer?

Originally Posted by rgthom
Another clear way of stating the situation.
Took me 14 pages to think of it.

5. peb
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## Re: Do you know the correct answer?

Originally Posted by CWSmith
As I did in post #24. Now, can you explain why one is right and one is wrong instead of just stating which you believe?
The table in post 24 is comically wrong, as has been pointed out multiple times in this thread. One only has to look at the first two columns to see that the number of times (rows) the contestants chooses the car is double what it should be. Yet they are treated as equal odds. Contestant chooses 1, car is behind 1 has two rows, car is behind 2 and behind 3 each have one row, same for the car being behind 2 and behind 3. Yet they are summed at the end as if that is all fine and dandy, its twice as likely the contestant chooses the car as it should be.

6. peb
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## Re: Do you know the correct answer?

Ok, one last attempt for the few malcontents who refuse to accept reality. We will consider 2 different card games, the first is not the same as the Monti-hall problem, the second is equivilent, and the difference should make the answer obvious to all.

Lets find the Ace of Spades, Version 1:

The host takes a deck of cards and shuffles. The contestant picks a card at random and leaves it face down. The Host takes a card at random and leaves it face down. Both cards have a 1/52 chance of winning. The remaining 50 cards are turned up and examined. Most times, of course, the ace of spades is in the remaining 50 cards, but you do it often enough and it won't be finally. In this case, it makes no difference if the contestant switches with the host or not. The odds are 50-50.

Lets Find the Ace of Spaces, Version 2.
The host takes a deck of cards and shuffles. The contestant picks a card at random and leaves it face down. The host then takes the remaining 51 cards, goes through them and find s the ace of spades and places it face down between him and the contestant (this is equivalent to Monti opening a goat every time, as Monti's rule is the same as "he must leave the car hidden"). He then hands the 50 cards to the player to verify there is no ace of spades in it and the contestant has a choice to switch. Of course he should switch, as there is a 51/52 chance the card the host turned face down is the ace of spades and only a 1/52 chance his card is the ace of spades.

The latter game is equivalent the the problem at hand, the former game yields the result the few obstinate folks want to believe is correct.

7. ## Re: Do you know the correct answer?

Originally Posted by George.
Choosing the door that Monty didn't open is identical to having chosen both the doors you didn't choose. You win if either one has the car, regardless of the fact that Monty shows you the one that doesn't. That's 66% odds.
That (speaking as an AI, which you have not, so far, adequatedly disproved) is cracking.

Andy

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## Re: Do you know the correct answer?

This is how the carnies at the county fair make a living.

9. ## Re: Do you know the correct answer?

Originally Posted by birlinn
Once the goat is revealed, there are now two doors; one will have a goat, one a car.
If you choose a door, there is a 50% chance it will have a car.
If you switch to the other door, there is a 50% chance it will have a car.
No gain in switching.
Not so.

We start with a bag of 3 seemingly identical objects, one of which is a valuable prize.

The probability of that bag containing the prize is 100%. The probability of any particular object in that bag, chosen at random is 33-1/3% (1/3).

The contestant selects an object from the back, at random. That object has a 33-1/3% probability of being the prize object.

Each of the remaining two objects in the bag have the same 33-1/3% probability of being the prize object, giving the bag a cumulative probability of 2/3 (66-2/3%).

Our preceptor, Monty Hall, knowing in advance, which object is the prize object and which objects are not, then draws a non-prize object from the bag and opens it, thus reducing the probability that that object was the prize to zero/zip/nada.

The bag, however, still has a 2/3 (66-2/3%) probability of containing the prize object. The 1/3 (33-1/3%) probability didn't just evaporate, it stayed with the bag, thus increasing the probability that the remaining object is the prize to 2/3 (66-2/3%).

This is Bayesian statistics at play. When the first object was selected by the contestant, it was even odds as to which object was the prize (we had no information).

Once Monty Hall reveals the non-prize object, we have introduced new information, and that changes the odds.

This is Bayesian statistics and inference at play:

Bayesian statistics is a system for describing epistemological uncertainty using the mathematical language of probability.
In the 'Bayesian paradigm,' degrees of belief in states of nature are specified; these are non-negative, and the total belief in all
states of nature is fixed to be one. Bayesian statistical methods start with existing 'prior' beliefs, and update these using data
to give 'posterior' beliefs, which may be used as the basis for inferential decisions.

http://www.scholarpedia.org/article/Bayesian_statistics

https://en.wikipedia.org/wiki/Bayesian_inference

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## Re: Do you know the correct answer?

Originally Posted by Tom Montgomery
OK... you've lost me. I do not understand this at all. Could you restate the above in a different way that I might be able to understand?
My brain is a bit fried. I've been working all day. However, your way of counting only combines realizations when the player selects the right door at the beginning. In effect, your count cheats the player by saying "It doesn't matter." However, it means that the option to stay with the first choice is undercounted in the win columns.

11. Old Guy
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## Re: Do you know the correct answer?

Originally Posted by Tom Montgomery
.
Let me ask you a question. Say the contestant chooses one door. Monty Hall keeps the other two doors. Monty Hall looks to see what are behind his two doors and opens a door containing a Goat and discards it. Say that no switch is offered. The remaining two doors are opened simultaneously. What is the probability that the Contestant has the car? What is the probability Monty Hall has the car? Are you under the impression they both have a 50% probability of having the Car?
Interesting. I'll need to think about that.

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## Re: Do you know the correct answer?

Originally Posted by George.
Choosing the door that Monty didn't open is identical to having chosen both the doors you didn't choose. You win if either one has the car, regardless of the fact that Monty shows you the one that doesn't. That's 66% odds.
Interesting way to think about it. I'm too tired to think clearly.

13. ## Re: Do you know the correct answer?

Mythbusters episode 177
Switching won

14. peb
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## Re: Do you know the correct answer?

Originally Posted by CWSmith
My brain is a bit fried. I've been working all day. However, your way of counting only combines realizations when the player selects the right door at the beginning. In effect, your count cheats the player by saying "It doesn't matter." However, it means that the option to stay with the first choice is undercounted in the win columns.
See post 553

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## Re: Do you know the correct answer?

Originally Posted by L.W. Baxter
the stunning thing is that this problem has been definitively solved. not here, out there. before this thread.
Um... yeah.

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## Re: Do you know the correct answer?

Originally Posted by peb
See post 553
I saw it. Other than a bit of attitude that I doubt you can back up, it didn't add anything.

17. ## Re: Do you know the correct answer?

In a moment of clarity, I realised the key to working out the odds is that if you switch, you will always reverse your original choice.

From there - original choice has 1:3 chance of being right, so switching has a 2:3 chance.

Simples

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## Re: Do you know the correct answer?

^^^^^^ this for me at #470 early this AM.
R

19. peb
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## Re: Do you know the correct answer?

You don't understand that you overcount correctly questing where the car is? You have to be trolling if that is the case.

20. peb
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## Re: Do you know the correct answer?

The idea that a 1 in 3 possibility of winning magically becomes 1 in 2 just because someone shows you one of the wrong answers is interesting concept.

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## Re: Do you know the correct answer?

Originally Posted by CWSmith
If that's true, then you should be able to create a table that proves it. I already did on the first page. Now you show your work.

Regarding your table in post 24:

Parenthetically, for simplicity you only need to show the possibilities for 'Contestant picks Door no. 1'

For that scenario, if the car is behind door no. 1, then you can't have can't have two rows for Monty's openings of doors 2 or 3, because they are reduced to the same scenario when he asks you if you wish to switch. As others have pointed out, most recently Peb, you are erroneously doubling the 'wins' if the contestant does not switch.

That is, the tables with three scenarios for 'Contestants pick door no. 1' are correct, and your table with four scenarios is incorrect.

Moreover, the tables with three scenarios are in line with the numerous narrative distillations that do not rely on tables, as well as the computer modelings.

Now, show me where the following (numbered) logical flow is in error:

i. I have a black bag with three marbles, one is made of diamond and the others glass.

ii. You pick a marble out of the bag and hold it in your closed hand without looking at it.

iii. I ask you if you would like to exchange your one marble for the two in my bag.

iv. Would you exchange your one marble for the two in my bag?

v. Of course you would. There is a 33.3% chance that you hold the diamond marble in your hand, but a 66.6% chance that I have the diamond in the bag.

vi. Alternatively, I remove the glass marble from my bag and show it to you, then toss it aside. Because, we both well know that one of the two marbles in my bag HAS to be made of glass.

vii. I then ask you, would you like to exchange your marble for the one in my bag? Of course you would, because by showing you the glass marble I have not magically changed the probability that the bag has a 66.6% chance of holding the diamond marble.

22. Originally Posted by peb
The idea that a 1 in 3 possibility of winning magically becomes 1 in 2 just because someone shows you one of the wrong answers is interesting concept.

That predilection is the basis for Three Card Monte, shell games, and more.

Kevin

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## Re: Do you know the correct answer?

Originally Posted by twodot
Now, show me where the following (numbered) logical flow is in error:
No. All you are doing is repeating what others have said and adding no understanding.

Find another monkey to dance for you.

24. Cease, cows
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## Re: Do you know the correct answer?

Originally Posted by CWSmith

Find another monkey to dance for you.

I was responding to a request by you...to me.

I responded.

And then I had a request for you.

Why not respond in kind?

25. Originally Posted by Breakaway
That predilection is the basis for Three Card Monte, shell games, and more.

Kevin

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I have often said that one shouldn't be allowed to graduate from high school without a solid grounding in probability and statistics.

More important than trig. More important calculus (even though you need calculus for more advance P+S)

One uses P+S every day of their life, knowingly or unknowingly -- is it safe to cross the street? Political polls, is buying a lottery ticket a good investment?

There are all sorts of things you don't consider as being probabilistic in nature.

26. ## Re: Do you know the correct answer?

Originally Posted by twodot
I had this on a t-shirt when I was in high school.

Didn’t turn out to be the chick-magnet that I thought it would.

27. ## Re: Do you know the correct answer?

Originally Posted by CWSmith
Okay, Nick, I'll see if this old dog can learn a new trick:

 Contestant Picks this Door Prize is Behind This Door Monty Shows This Door Stays Switches 1 1 2 Win Lose 1 1 3 Win Lose 1 2 3 Lose Win 1 3 2 Lose Win 2 1 3 Lose Win 2 2 1 Win Lose 2 2 3 Win Lose 2 3 1 Lose Win 3 1 2 Lose Win 3 2 1 Lose Win 3 3 1 Win Lose 3 3 2 Win Lose Win % 0.5 0.5

Okay, Tom, I told you what I thought was wrong with your chart. Now I'd like you to tell me what is wrong with mine.

Edit - Sorry. I had a typo that I corrected.

I think the issue here is that the table treats each goat uniquely.

We’ll call them Lee and Paul.

So when the contestant chooses the correct door, you handled the outcomes of exposing Lee and exposing Paul as unique events with their own probability.

But the game deals in combinations, not permutations, so in goat form, Lee and Paul are indiscernible, and the exposing of either can be considered a singular event.

This would eliminate three rows from your table, all of which favored keeping your original choice.

28. Old Guy
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## Re: Do you know the correct answer?

Originally Posted by twodot
I was responding to a request by you...to me.

I responded.

And then I had a request for you.

Why not respond in kind?
I have responded to that problem already.

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## Re: Do you know the correct answer?

Originally Posted by Beowolf
I think the issue here is that the table treats each goat uniquely.

We’ll call them Lee and Paul.

So when the contestant chooses the correct door, you handled the outcomes of exposing Lee and exposing Paul as unique events with their own probability.

But the game deals in combinations, not permutations, so in goat form, Lee and Paul are indiscernible, and the exposing of either can be considered a singular event.

This would eliminate three rows from your table, all of which favored keeping your original choice.
I think I know the rows you want to remove so that it looks like the earlier table, but the point is there really are two goats behind two doors.

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## Re: Do you know the correct answer?

Everyone is just repeating the same arguments and that includes me.

Until someone has a new idea, how about we let this go? Otherwise, we can just number the ideas we have (I think there are two and all the rest are just rephrasings of those two) and throw those numbers instead of doing all this typing.

31. ## Re: Do you know the correct answer?

meh-eh-eh-eh

put a simpler way: there are repeat numbers in the "prize behind this door" column, giving the contestant an extra win staying with each door, and an extra loss switching.

32. ## Re: Do you know the correct answer?

Originally Posted by CWSmith
Everyone is just repeating the same arguments and that includes me.

Until someone has a new idea, how about we let this go? Otherwise, we can just number the ideas we have (I think there are two and all the rest are just rephrasings of those two) and throw those numbers instead of doing all this typing.
What numbers? 1 for switch and 2 in 3 odds and 2 for stay and 1 in 3 odds and 3 for 50/50 odds?
Last edited by Jimmy W; 06-14-2022 at 12:05 AM.

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## Re: Do you know the correct answer?

Originally Posted by CWSmith
I have responded to that problem already.

Yes, but I responded to your request, directed to me.

And that was my request, directed to you.

If you wish to pull rank on me, how so?

34. ## Re: Do you know the correct answer?

Originally Posted by CWSmith
I think I know the rows you want to remove so that it looks like the earlier table, but the point is there really are two goats behind two doors.
Yes, but key in this instance is that 2 times out of three when a goat is removed from play by opening a door, the other door is the car. One time out of three the other door is also a goat.

The initial choice by the contestant is meaningless other than triggering the host. It’s truely random. The host has the only real choice, but is stymied two times out of three into revealing the car by the rules of the game (he must open a door and it must be a goat). The contestant then must always choose to switch to benefit. He will only lose when the host has two goats behind two doors, which will only happen 1/3 of the time with a random shuffle.

If he doesn’t switch, he will win only 1/3 of the time because 2/3 of the time the host has one goat behind his two doors, and the car. And he must reveal the goat, and so points a finger at the car. 1/3 of the time he lies and point a finger at a goat because he has two.

Scotch and a good sleep worked for me
Last edited by Decourcy; 06-14-2022 at 12:13 AM.

35. ## Re: Do you know the correct answer?

To me, it makes more sense to think in terms of a game like chess. The first move isn’t designed to win, but to provoke a move that puts the pieces where you want them.

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